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2 years ago

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An object of mass m is dropped from height h above a planet of mass M and radius R.Find an expression for the object's speed as

it hits the ground

1 answer:

Mnenie [13.5K]2 years ago

4 0

Answer:

Explanation:

Mass of object = m

Height above planet = h

Mass of planet = M

Radius of planet = R

As we have to find out velocity, so let's apply the law of conservation of energies on initial( when the object was at height) and final( when object hit the surface points.

Initial energy = Final energy

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2}mv_{i} ^{2} - \frac{GMm}{h+R} = \frac{1}{2}mv_{f} ^{2} - \frac{GMm}{R}\\v_{i} = 0\\v_{f} ^{2} = 2\frac{GM}{R} - 2 \frac{GM}{h+R}$

$v_{f} =\sqrt{\frac{2GMh}{R(R+h)} }$

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

$P = \ro g h = 1000 * 10 * 20 = 200000 Pa$

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

$P_1V_1 = P_2V_2$

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

$V_1 = Ah$

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

$P_1Ah_1 = P_2Ah_2$

$h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m$

b) If the temperatures changes, we can still reuse the ideal gas equation above:

$\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}$

$h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m$

3 0

2 years ago

The bird is held in level flight due to the force exerted on it by the air as the bird beats its wings. What is the maximum valu

Cerrena [4.2K]

Answer:

maximumforce is F = mg

Explanation:

For this case we must use Newton's second law,

Σ F = m a

bold indicate vectors, so we will write it in its components x and y

X axis

Fₓ = maₓ

Axis y

Fy - W = m a $a_{y}$

Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components

Cos θ = Fₓ / F

Fₓ = F cos θ

sin θ = Fy / F

Fy = F sin θ

Let's replace and calculate

F sin θ -w = m a

As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)

F sin θ = w = mg

The maximum value of this equation occurs when the sin=1, in this case

F = mg

3 0

2 years ago

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

1 year ago

Read 2 more answers

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is $\tau = 48t \r k$

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago

Which of the following are dwarf planets? Check all that apply. Ceres Namaka Eris Charon Haumea Makemake Pluto

kicyunya [14]

Ceres: Yes!
Namaka: No!
Eris: Yes!
Charon: No. (it's a satellite, and dwarf planet's can't be satellites!)
Haumea: Yes!
Makemake: Yes!
Pluto: Yes!

Glad To Help;)

6 0

1 year ago

Read 2 more answers